Consider a topological space $(X,\mathcal{T})$, and a sequence of sets $(K_n)_{n\in\omega}$, such that $(\forall n\in\omega)(K_n\text{ is compact})$, and $(\forall m<n)(K_m\supset K_n)$. Is it true that $\cap_{n\in\omega}K_n\neq\emptyset$?


1 Answer 1


As pointed out elsewhere you need to assume the $K_n$ are all non-empty. However, that assumption is not strong enough: under the cofinite topology on $\Bbb{R}$, every subset of $\Bbb{R}$ is compact. So if you take $K_n = \{n, n + 1, n + 2, \ldots\}$, you get a counter-example to your claim.

If you assume $(X, {\cal T})\,$ is Hausdorff, your claim holds. (If $\bigcap_n K_n\,$ were empty, the sets $K_1 \setminus K_n\,$ would be open, because of Hausdorffness, and would cover $K_1$. Hence a finite subset of these open sets would cover $K_1$, implying that the $K_n$ are eventually all empty).

All the above argument relies on is that the $K_i$ are all closed. So it works if we take that as an assumption or if we assume that $X$ is a $KC$-space, one in which all compact subsets are closed.

  • $\begingroup$ We can use a weaker assumption: topology.pi-base.org/properties/P000100 For example, the one point compactification of the rationals has this property but isn't T_2. $\endgroup$ Sep 25 at 23:06
  • $\begingroup$ @StevenClontz: thanks for pointing that out. $\endgroup$
    – Rob Arthan
    Sep 25 at 23:07
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    $\begingroup$ @RobArthan the point is, if $K_m = \{\}$ then you can't possibly have $K_m \supsetneq K_n$. $\endgroup$ Sep 26 at 8:31
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    $\begingroup$ That's a good point, but the result is stronger to assume $\subset$ allows equality, with the caveat that sets are nonempty. $\endgroup$ Sep 26 at 11:16
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    $\begingroup$ Perhaps you should mention in your answer that spaces in which all compact sets are closed are known as KC-spaces. And of course the claim is true if all $K_n$ are assumed to be closed ... $\endgroup$
    – Paul Frost
    Sep 29 at 15:44

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